Verify that Pfaffian differential equation yzdx +($x^{2}$y-zx)dy +($x^{2}$z-xy)dz is integrable and find the corresponding integral.

Verify that Pfaffian differential equation yzdx +($x^{2}$y-zx)dy +($x^{2}$z-xy)dz=0 is integrable/exact and find the corresponding integral.

Solution: The given Pfaffian Differential equation are

                yzdx +($x^{2}$y-zx)dy +($x^{2}$z-xy)dz=0  \rightarrow(i)

Comparing with Pdx+Qdy+Rdz =0 , we get

P=yz , Q = $x^{2}$y-zx , R= $x^{2}$z-xy

$\therefore \bar{X}$  = (yz , $x^{2}$y-zx , $x^{2}$z-xy)

$\therefore Curl\;\overline X\;=$

$\begin{bmatrix}\widehat i&\widehat j&\widehat k\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\yz&x^2y-zx&x^2z-xy\end{bmatrix}$

$\begin{array}{l}=-2\widehat j(xz-y)-\;2\widehat k(xy-z)\\=(0,-2(xz-y),-\;2(xy-z))\end{array}$

Clearly, $\overline X\;.Curl\;\overline X\;=\;0$

Therefore, the given Pfaffian differential equation is integrable.

Suppose x is treated as constant, then dx=0.

Then equation (i) becomes,

  ($x^{2}$y-zx)dy +($x^{2}$z-xy)dz=0 

$ \Rightarrow (xy-z)dy + (xz-y) = 0$

$\Rightarrow x(ydy+zdz) - (zdy + ydz) = 0$

Integrating we get

$x(\frac{y^2}2\;+\;\frac{z^2}2)\;-\;yz\;=c_1$

Let U(x,y,z) = $x(\frac{y^2}2\;+\;\frac{z^2}2)\;-\;yz\;$

$\begin{array}{l}\begin{array}{l}\therefore\mu=\;\frac1Q.\frac{\partial U}{\partial y}\\\;\;\;\;\;=\frac1{x^2y-zx}.(xy-z)\\\;\;\;\;\;=\frac1{x(xy-z)}.(xy-z)\\\;\;\;\;\;=\frac1x\\\therefore K=\mu P\;-\;\frac{\partial U}{\partial x}\end{array}\\\;\;\;\;\;=\frac1x.yz-\frac12.(y^2+z^2)\\\;\;\;\;\;=\frac{yz-{\displaystyle\frac12}x(y^2+z^2)}x\\\;\;\;\;\;=-\frac Ux\end{array}$

$\begin{array}{l}\begin{array}{l}\frac{dU}{dx}+\;K(U,x)=0\\\Rightarrow\frac{\displaystyle dU}{\displaystyle dx}-\frac Ux=0\\\Rightarrow\frac{\displaystyle dU}{\displaystyle U}-\frac{dx}x=0\\\Rightarrow\frac Ux=c\\\Rightarrow U=cx\end{array}\\\Rightarrow x(\frac{y^2}2\;+\;\frac{z^2}2)\;-\;yz\;=cx\\\Rightarrow x(y^2+z^2)\;-\;2yz\;=2cx\\\end{array}$